Question

The ratio of the wavelengths of the light absorbed by a Hydrogen atom when it undergoes n = 2 $\rightarrow$ n = 3 and n = 4 $\rightarrow$ n = 6 transitions, respectively, is:

• A. 1 : 9
• B. 1 : 36
• C. 1 : 4
• D. 1 : 25

Hint:

For hydrogen atom transitions, the wavelength is inversely proportional to the energy difference, which can be calculated using the Rydberg formula. The energy difference determines the wavelength of light absorbed or emitted.

Answer 1

The Correct Option is C

The energy difference (\(\Delta E\)) for a hydrogen atom transition, absorbed or emitted light's wavelength (\(\lambda\)), is defined by \(\Delta E = \frac{R_H}{n_1^2} - \frac{R_H}{n_2^2}\), where \(R_H\) is the Rydberg constant and \(n_1, n_2\) are the principal quantum numbers of the initial and final orbits. Energy and wavelength are inversely proportional: \(E = \frac{hc}{\lambda}\), with \(h\) as Planck's constant and \(c\) as the speed of light. Consequently, wavelength ratios are the inverse of energy ratios. Calculations for energy differences of two transitions: 1. For \(n = 2 \rightarrow n = 3\): \[\Delta E_1 = \frac{R_H}{2^2} - \frac{R_H}{3^2} = R_H \left( \frac{1}{4} - \frac{1}{9} \right) = R_H \times \frac{5}{36}\] 2. For \(n = 4 \rightarrow n = 6\): \[\Delta E_2 = \frac{R_H}{4^2} - \frac{R_H}{6^2} = R_H \left( \frac{1}{16} - \frac{1}{36} \right) = R_H \times \frac{5}{144}\] The ratio of wavelengths (\(\lambda_1 / \lambda_2\)) equals the inverse ratio of energy differences: \[\frac{\lambda_1}{\lambda_2} = \frac{\Delta E_2}{\Delta E_1} = \frac{\frac{5}{144}}{\frac{5}{36}} = \frac{36}{144} = \frac{1}{4}\] Thus, the ratio of wavelengths is 1 : 4. Therefore, the correct answer is (3) 1 : 4.